A box contains 28 bulbs of which 7 bulbs are defective a bulb is drawn randomly from the box . Find the probability of getting a non defective bulb
Answers
Answered by
3
Answer:
total bulbs=28
defective bulbs =7
non defective bulbs=21
p(e) 21/28 =3/4 =0.75
Answered by
3
Answer:
3/4 is the correct answer.
Step-by-step explanation:
n(S) :- 28
Let, event A is that to get a non-defevtive bulb.
n(A) :- 28-7 = 21
p(A) = n(A) / n(S)
= 21 / 28
= 3/4
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